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Binomial Distribution Question

  1. Oct 3, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    A good hitter in baseball has a batting average of .300 which means that the hitter will be successful three times out of 10 tries on average. Assume that the batter has four times at bat per game.


    a) What is the probability that he will get two hits or less in a three game series?
    b) What is the probability that he will get five or more hits in a three-game series?

    2. Relevant equations

    Binomial distribution: [tex] P_n(n) = \frac{N!}{n!(N-n)!}p^nq^{(N-n)}[/tex]

    3. The attempt at a solution


    For a) I use
    p = 0.300
    q = 0.700
    n = 2 hits or less
    N = 12 aka total at bats

    I solve directly and get an answer of 0.168.

    This seems reasonable to me.

    Am I using the binomial distribution correctly here? If so, I will proceed to part b.

    Thank you.
     
  2. jcsd
  3. Oct 3, 2016 #2

    PeroK

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    Is that for exactly two hits or two hits or fewer?

    Did you know that the Excel spreadsheet has a BINOMDIST function?
     
  4. Oct 3, 2016 #3

    RJLiberator

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    This is for Two hits or less in 12 at bats.

    I was worried about that issue. Would I need to calculate P_12(1) and P_12(0) and add the probabilities together to get the correct answer?
     
  5. Oct 3, 2016 #4

    PeroK

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    Yes, the formula you quoted is for exactly ##n## hits in ##N## trials. To get ##n## or fewer you need the sum of that formula from ##0## to ##n##.

    If you have Excel check it out, as you can do "cumulative" = "True" or "False" as required:

    =BINOMDIST(2,12,0.3,TRUE)
    =BINOMDIST(2,12,0.3,FALSE)

    Will give the value for two hits or fewer and two hits exactly respectively.
     
  6. Oct 3, 2016 #5

    RJLiberator

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    Allright, so now the answer to A is P_12(2)+P_12(1)+P_12(0) = 0.168+0.014+0.071 = 0.253

    This makes more reasonable sense to me and I see how it makes sense with the additions of the probabilities as it relates to the question. Now I will consider part b.
     
  7. Oct 3, 2016 #6

    RJLiberator

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    Now that I got the hang of this, this is quite easy (I think).

    So for part b, all I did was calculate the probabilities P_12(4) and P_12(3). I used my previous answer in A and summed up these values. I then took 1-this sum and that is the probability that he gets 5 hits or more.

    This turned out to be 0.276 which is a reasonable number.
     
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